Score on last try: 0 of 1 pts. See Details for more. You can retry this question below Find the final amount of money in an account if \( \$ 3,100 \) is deposited at \( 7 \% \) interest compounded weekl and the money is left for 6 years. The final amount is \( \$ \square \) Round answer to 2 decimal places Submit Question

To find the final amount of money in an account after depositing $3,100 at a 7% interest compounded weekly for 6 years, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{n} \right)^{nt} \] where: - \( A \) is the final amount of money in the account - \( P \) is the principal amount (initial deposit) = $3,100 - \( r \) is the annual interest rate = 7% = 0.07 - \( n \) is the number of times interest is compounded per year = 52 (since it is compounded weekly) - \( t \) is the time in years = 6 years Substitute the given values into the formula: \[ A = 3100 \left( 1 + \frac{0.07}{52} \right)^{52 \times 6} \] Now, we can calculate the final amount of money in the account. Calculate the value by following steps: - step0: Calculate: \(3100\left(1+\frac{0.07}{52}\right)^{52\times 6}\) - step1: Divide the terms: \(3100\left(1+\frac{7}{5200}\right)^{52\times 6}\) - step2: Add the numbers: \(3100\left(\frac{5207}{5200}\right)^{52\times 6}\) - step3: Multiply the numbers: \(3100\left(\frac{5207}{5200}\right)^{312}\) - step4: Simplify: \(3100\times \frac{5207^{312}}{5200^{312}}\) - step5: Rewrite the expression: \(100\times 31\times \frac{5207^{312}}{5200^{312}}\) - step6: Rewrite the expression: \(100\times 31\times \frac{5207^{312}}{100^{312}\times 52^{312}}\) - step7: Reduce the numbers: \(31\times \frac{5207^{312}}{100^{311}\times 52^{312}}\) - step8: Multiply the fractions: \(\frac{31\times 5207^{312}}{100^{311}\times 52^{312}}\) The final amount of money in the account after depositing $3,100 at a 7% interest compounded weekly for 6 years is approximately $4716.75. Therefore, the final amount is $4716.75.